Double Integral MCQ Quiz - Objective Question with Answer for Double Integral - Download Free PDF

Explanation:

 ${\int }_{x=3}^{x=4}{\int }_{y=1}^{y=2}\frac{dxdy}{(x+y{)}^2}$

= ${\int }_{x=3}^{x=4}-{\left[\frac{1}{(x+y)}dx\right]}_{y=1}^{y=2}$

$=-{\int }_{x=3}^{x=4}[\frac{1}{2+x}-\frac{1}{x+1}]$

= $-[\ln (2+x)-\ln (x+1){]}_{x=3}^{x=4}$

= $-{[\ln \left(\frac{2+x}{x+1}\right)]}_3^4$

$=-[\ln \left(\frac{6}{5}\right)-\ln \left(\frac{5}{4}\right)]$

= $-[\ln \left(\frac{6\times 4}{5\times 5}\right)]$

$=-\ln \left(\frac{24}{25}\right)$

$=\ln \left(\frac{25}{24}\right)$

Thus, ${\log }_e\left(\frac{a}{24}\right)=\ln \left(\frac{25}{24}\right)$

∴ a = 25

Explanation:

Given curve, ${\mathrm{y}}^2={\mathrm{x}}^2\left(4^2-{\mathrm{x}}^2\right)\Rightarrow \mathrm{y}=\pm \mathrm{x}\sqrt{4^2-{\mathrm{x}}^2}$

At x = 0 ⇒ y = 0

$\begin{array}{l}\mathrm{x}=4\Rightarrow \mathrm{y}=0\\ \mathrm{x}=-4\Rightarrow \mathrm{y}=0\end{array}$

So the diagram for the given curve is

Required area $\mathrm{A}=\int \mathrm{y}.\mathrm{d}\mathrm{x}$

$\begin{array}{l}=2\times \underset{0}{\overset{4}{\int }}\left({\mathrm{y}}_1-{\mathrm{y}}_2\right).\mathrm{d}\mathrm{x}\\ =2\underset{0}{\overset{4}{\int }}2\mathrm{x}\sqrt{16-{\mathrm{x}}^2}.\mathrm{d}\mathrm{x}\\ =4\underset{0}{\overset{4}{\int }}\mathrm{x}\sqrt{16-{\mathrm{x}}^2}.\mathrm{d}\mathrm{x}\end{array}$

Put $\mathrm{x}=4\mathrm{s}\mathrm{i}\mathrm{n}\theta \Rightarrow \mathrm{d}\mathrm{x}=4\mathrm{C}\mathrm{o}\mathrm{s}\theta .\mathrm{d}\theta$

$\begin{array}{l}\mathrm{A}=4\underset{\theta =0^{\circ }}{\overset{\theta ={90}^{\circ }}{\int }}4\mathrm{S}\mathrm{i}\mathrm{n}\theta 4\mathrm{C}\mathrm{o}\mathrm{s}\theta .4\mathrm{c}\mathrm{o}\mathrm{s}\theta \mathrm{d}\theta \\ \mathrm{A}=4^4\underset{\theta =0^{\circ }}{\overset{\theta ={90}^{\circ }}{\int }}\mathrm{S}\mathrm{i}\mathrm{n}\theta \mathrm{C}\mathrm{o}{\mathrm{s}}^2\theta .\mathrm{d}\theta \\ \mathrm{A}=256\times -{\left(\frac{\mathrm{C}\mathrm{o}{\mathrm{s}}^3\theta }{3}\right)}_{0^{\circ }}^{{90}^{\circ }}\\ =-\frac{256}{3}\left[0-1\right]=\frac{256}{3}\end{array}$

$\frac{3A}{16}=\frac{3}{16}\cdot \frac{256}{3}$ = 16 

Hence Option(2) is the correct answer.

Concept:

Surface area of portion of plane in cylinder is:

S= ${\iint }_C\sqrt{1+{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2}dA$

Explanation: 

Given y + 2z = 2 and $x^2+y^2=5$

$\Rightarrow z=\frac{1}{2}(2-y)$

$\Rightarrow \frac{\partial z}{\partial x}=0and\frac{\partial z}{\partial y}=-\frac{1}{2}$

Therefore Surface area of portion of plane in cylinder is:

S= ${\iint }_C\sqrt{1+{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2}dA$

$={\iint }_C\sqrt{1+0+{(-\frac{1}{2})}^2}dA$

$={\iint }_C\frac{\sqrt{5}}{2}dA=\frac{\sqrt{5}}{2}{\iint }_{C}dA$

$=\frac{\sqrt{5}}{2}\times 5\pi =\frac{5\sqrt{5}\pi }{2}$

Hence Option(2) is the correct answer.

Explanation:

${\int }_c(xyi+y^{2}j).(idx+jdy)$

= ${\int }_c(xydx+y^{2}dy)$

= ${\int }_{-1}^1{\int }_{-1}^1(0-x)dxdy$ (as ${\int }_c(Pdx+Qdy)=\int {\int }_S(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy$)

= 0 (as ${\int }_{-a}^{a}f(x)dx=0$ if f(x) is odd function)

Option (3) is true.

Calculation:

Given, f(x, y) = ${\displaystyle {\iint }_{(u-x{)}^2+(v-y{)}^2\le 1}}$ $e^{-\sqrt{(u-x{)}^2+(v-y{)}^2}}$ du dv.

Let u - x = r cosθ and v - y = r sinθ 

⇒ du dv = r dr dθ

and, (u - x)² + (v - y)² ≤ 1 represents a circle of radius, r ≤ 1.

∴  ${\displaystyle {\iint }_{(u-x{)}^2+(v-y{)}^2\le 1}}$ $e^{-\sqrt{(u-x{)}^2+(v-y{)}^2}}$ du dv

= ${\displaystyle {\int }_0^1{\int }_0^{2\pi }r{e}^{-r}drd\theta }$

= ${\displaystyle {\int }_0^{1}r{e}^{-r}[\theta {]}_0^{2\pi }dr}$

=2π${\displaystyle {\int }_0^{1}r{e}^{-r}dr}$

= 2π${\displaystyle [r\int e^{-r}dr-\int (\frac{d}{dr}r\int e^{-r}dr)dr]}$

= 2π${\displaystyle {[-r{e}^{-r}-e^{-r}]}_0^1}$

= 2π[(- e^-1 - e^-1) - (0 - 1)]

= 2π(1 − 2e−1) 

∴  ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}}$ f(n, n2) = 2π(1 − 2e−1) 

(4) is correct

Given:

$I={\int }_{x=0}^1{\int }_{y=0}^{x^2}x{y}^{2}dydx$

0 ≤ y ≤ x² (this is represented by vertical strip)

And x varies from 0 to 1.

Now if we change the order of integration, we have to draw a horizon strip.

After changing the order of Integration

$\sqrt{y}\le x\le 1$

And, 0 ≤ y ≤ 1

∴ $I={\int }_{y=0}^1{\int }_{x=\sqrt{y}}^{1}x{y}^{2}dxdy$

$\underset{0}{\overset{2}{\int }}\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{x}+\mathrm{y}}\mathrm{d}\mathrm{y}\mathrm{d}\mathrm{x}=\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{\mathrm{x}}\left(\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{y}}\mathrm{d}\mathrm{y}\right)\mathrm{d}\mathrm{x}$

$=\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{\mathrm{x}}{\left({\mathrm{e}}^{\mathrm{y}}\right)}_0^{\mathrm{x}}\mathrm{d}\mathrm{x}=\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{\mathrm{x}}\left({\mathrm{e}}^{\mathrm{x}}-1\right)\mathrm{d}\mathrm{x}$

$=\underset{0}{\overset{2}{\int }}\left({\mathrm{e}}^{2\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\right)\mathrm{d}\mathrm{x}={\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}-{\mathrm{e}}^{\mathrm{x}}\right)}_0^2$

$=\frac{{\mathrm{e}}^4}{2}-{\mathrm{e}}^2-\frac{1}{2}+1=\frac{{\mathrm{e}}^4}{2}-{\mathrm{e}}^2+\frac{1}{2}$

$=\frac{1}{2}\left({\mathrm{e}}^4-2{\mathrm{e}}^2+1\right)=\frac{1}{2}{\left({\mathrm{e}}^2-1\right)}^2$

The equation of line in intercept form is given by

$\frac{x}{2}+\frac{y}{1}=1$

⇒ $y=1-\frac{x}{2}\Rightarrow x=2-2y$

⇒ $\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2-2y}{\int }}\left(xydx\right)dy=\underset{0}{\overset{1}{\int }}\left[\left(\frac{y{x}^2}{2}\right){|}_0^{2-2y}\right]dy$

$=\underset{0}{\overset{1}{\int }}\frac{y}{2}{\left(2-2y\right)}^{2}dy$

$=\underset{0}{\overset{1}{\int }}2y{\left(1-y\right)}^{2}dy=\underset{0}{\overset{1}{\int }}(2y-4y^2+2y^3)dy$

$={\left[y^2-\frac{4}{3}{y}^3+\frac{y^4}{2}\right]}_0^1=\frac{1}{6}$

Given,

$\frac{1}{2\pi }\int \underset{D}{\overset{}{\int }}\left(x+y+10\right)dxdy$

and x² + y² ≤ 4

Putting x = r.cosθ, y = r.sinθ and dx.dy = r.dr.dθ

$\mathrm{I}=\frac{1}{2\pi }\underset{\theta =0}{\overset{2\pi }{\int }}\underset{\mathrm{r}=0}{\overset{2}{\int }}\left(\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{s}\theta +\mathrm{r}\mathrm{s}\mathrm{i}\mathrm{n}\theta +10\right)\mathrm{r}\mathrm{d}\mathrm{r}\mathrm{d}\theta$

$=\frac{1}{2\pi }\underset{\theta =0}{\overset{2\pi }{\int }}{\left(\frac{{\mathrm{r}}^3}{3}\mathrm{c}\mathrm{o}\mathrm{s}\theta +\frac{{\mathrm{r}}^3}{3}\mathrm{s}\mathrm{i}\mathrm{n}\theta +5{\mathrm{r}}^2\right)}_0^2\mathrm{d}\theta$

$=\frac{1}{2\pi }\left[\underset{\theta =0}{\overset{2\pi }{\int }}\left(\frac{8}{3}\mathrm{c}\mathrm{o}\mathrm{s}\theta +\frac{8}{3}\mathrm{s}\mathrm{i}\mathrm{n}\theta +20\right)\mathrm{d}\theta \right]$

$=\frac{1}{2\pi }\left[0+0+20\left(2\pi \right)\right]=20$

The given curves are y = x and y = x²

Solving the equations, we get

x  = 0, x = 1

$Area=\underset{0}{\overset{1}{\int }}\left(x-x^2\right)dx$

$={\left(\frac{x^2}{2}-\frac{x^3}{3}\right)}_0^1=\frac{1}{2}-\frac{1}{3}$

$=\frac{1}{6}squnits$

Alternate Solution:

Concept:

Area of a region can be calculated by:

$\int \int dxdy$

Calculation:

Solving equation y = x² and we get y = x

We get intersection points i.e. (0,0) and (1,1)

Area of the region:

$\underset{x=0}{\overset{x=1}{\int }}\underset{y=x^2}{\overset{y=x}{\int }}dy.dx$

$\underset{x=0}{\overset{x=1}{\int }}{\left[y\right]}_{x^2}^x.dx$

$\underset{x=0}{\overset{x=1}{\int }}\left(x-x^2\right)dx$

${\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

$Given:\underset{y=0}{\overset{\pi }{\int }}\underset{x=y}{\overset{\pi }{\int }}\frac{\sin x}{x}dxdy$

Changing the order and take vertical strip, we get:

$I=\underset{x=0}{\overset{\pi }{\int }}\underset{y=0}{\overset{y=x}{\int }}\frac{\sin x}{x}dxdy$

$\underset{x=0}{\overset{\pi }{\int }}\frac{\sin x}{x}dx\underset{y=0}{\overset{y=x}{\int }}dy$

$\Rightarrow \underset{x=0}{\overset{\pi }{\int }}\frac{\sin x}{x}dx{\left[y\right]}_0^x$

$\Rightarrow \underset{x=0}{\overset{\pi }{\int }}\frac{\sin x}{x}dx\left[x-0\right]$

$\Rightarrow I=\underset{x=0}{\overset{\pi }{\int }}\sin xdx$

$I=(-cosx{)}_0^{\pi }$

I = (1 – (-1))

I = 2

Explanation:

We have the integration given as,

$I=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{x^2}{\int }}\left(x^2+y^2\right)dydx$

$I=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{x^2}{\int }}\left(x^2+y^2\right)\cdot dy\cdot dx$

$I=\underset{0}{\overset{1}{\int }}{\left[x^{2}y+\frac{y^3}{3}\right]}_0^{x^2}dx$

$I=\underset{0}{\overset{1}{\int }}\left[x^4+\frac{x^6}{3}\right]dx$

Placing the limits we get,

$I={\left[\frac{x^5}{5}+\frac{x^7}{7\times 3}\right]}_0^1$

$I=\frac{1}{5}+\frac{1}{21}=\frac{26}{105}$

Hence the required value of integration will be $\frac{26}{105}$.

Concept:

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Length major axis of ellipse = 2a

Length of minor axis of ellipse = 2b

Area $=\int \underset{R}{\overset{}{\int }}1\cdot dy\cdot dx$

Calculation:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow y=\sqrt{b\left(1-\frac{x^2}{a^2}\right)}$.

For the first quadrant, take a vertical strip as shown. Here, y coordinate varies from 0 to $\sqrt{b\left(1-\frac{x^2}{a^2}\right)}=\frac{b}{a}\sqrt{a^2-x^2}$.

Also, the x-coordinate varies from 0 to a

∴ Area $=\underset{0}{\overset{a}{\int }}\underset{0}{\overset{\frac{b}{a}\sqrt{a^2-x^2}}{\int }}\left(1\right)dydx$

$=\underset{0}{\overset{a}{\int }}\left(\frac{b}{a}\sqrt{a^2-x^2}\right)dx$

$=\frac{b}{a}{\left[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)\right]}_0^a=\frac{b}{a}\times \frac{a^2}{2}\times \frac{\pi }{2}=\frac{\pi ab}{4}$

∴ The total area of ellipse $=4\times \frac{\pi ab}{4}$ = πab units

Important Points

1) x² + y² = a² ; Represents a circle centred at (0, 0) and radius ‘a’ units.

2) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$; Represents Ellipse with major axis length 2a and minor axis length 2b and vertex at (0,0)

3) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$; Equation of Hyperbola .

Calculation

Given equations are: y² = 9x, x – y + 2 = 0

By solving the above two equations,

The point of intersection of the two curves are: (1, 3) and (4, 6)

Now, the graph is shown below.

By considering the horizontal strip,

The limits of y are:  3 to 6

The limits of x are: (y – 2) to y²/9

Now, the required area is $=\int \int dxdy$

$=\underset{3}{\overset{6}{\int }}\underset{y-2}{\overset{\frac{y^2}{9}}{\int }}dxdy$

$=\underset{3}{\overset{6}{\int }}\left(\frac{y^2}{9}-y+2\right)dy$

$={\left[\frac{y^3}{27}-\frac{y^2}{2}+2y\right]}_3^6=\frac{1}{2}$

Concept:

Gauss divergence theorem

$\int \underset{S}{\overset{}{\int }}F\cdot nds=\underset{S}{\overset{}{\int }}div\cdot Fdv$ 

Calculation:

$\int \underset{S}{\overset{}{\int }}F\cdot nds=\underset{S}{\overset{}{\int }}div\cdot Fdv$

$=\underset{S}{\overset{}{\int }}\mathrm{\nabla }\left[\left(x+y\right)\hat{i}+\left(x+z\right)\hat{j}+\left(y+z\right)\hat{k}\right]dv$

$=\underset{S}{\overset{}{\int }}\left(1+0+1\right)dv$

$=2\int dv=2V$

$V=\frac{4}{3}\pi {\left(3\right)}^3=36\pi$

2 × 36 π = 72 π = 226.19